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Lemma 29: Suppose is a finite set of infinite degree elvations and is compact. Then for all sufficiently large , there exists an intermediate covering such that

(a) embeds in

(b) every descends to an elevation of degree

(c) the are pairwise distinct

Proof: We claim that the images of never share an infinite ray (a ray is an isometric embedding of ). Neither do two ends of the same elevation. Let’s claim by passing to the universal cover of, a tree .

For each , lift to a map . If and share an infinite ray then there exists such that and overlay in an infinite ray. The point is that correspond to cosets and . But this implies that

This implies that . So . A similar argument implies that the two ends of do not overlap in an infinite ray. This proves the claim.

Let be the core of . Enlarging if necessary, we can assume that

(i) ;

(ii) is a connected subgraph;

(iii) for each , for some , ;

(iv) for each , .

For each identifying with so that is identified with and is identified with . Let

For all sufficiently large ,

Now, the restriction of factors through , where . This is a finite-to-one immersion, so, by theorem 5, we can complete it to a finite-sheeted covering map as required.

Theorem 19: is LERF.

Recall the set-up from the previous lecture. We built a graph of spaces for .

Proof: Let be finitely generated. Let be the corresponding covering space of and let be compact. Because is finitely generated, there exists a subgraph of spaces such that . We can take large enough so that . We can enlarge so that it contains every finite-degree edge space of . Also enlarge so that

for any . For each let and let incident edge map of infinite degree .

Applying lemma 29 to , for some large , set . (Here we use the fact that vertex groups of are finitely generated)

Define as follows:

For each , the edge space is the that the lemma produced from the corresponding .

Now, by construction, can be completed to a graph of spaces so that the map

factors through and embeds. Let be identical to except with +’s and -‘s exchanged. Clearly satisfies Stallings condition, as required.